Question: Let $h$ be a polynomial function and let $h'$, its derivative, be defined as $h'(x)=x^2(x-1)(x+2)$. At how many points does the graph of $h$ have a relative minimum ? Choose 1 answer: Choose 1 answer: (Choice A) A None (Choice B) B One (Choice C) C Two (Choice D) D Three
Explanation: We can find the relative extrema (i.e. minima and maxima) of $h$ by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. We are given that $h'(x)=x^2(x-1)(x+2)$. $h'(x)=0$ for $x=-2,0,1$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=-2$, $x=0$, and $x=1$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into four intervals: $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $(-\infty, \llap{-}2)$ $( \ \ \llap{-}2,0)$ $(0,1)$ $(1,\infty)$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $(-\infty,-2)$ $x=-3$ $h'(-3)=36>0$ $h$ is increasing $\nearrow$ $(-2,0)$ $x=-1$ $h'(-1)=-2<0$ $h$ is decreasing $\searrow$ $(0,1)$ $x=\dfrac{1}{2}$ $h'\left(\dfrac12\right)=-\dfrac{5}{16}<0$ $h$ is decreasing $\searrow$ $(1,\infty)$ $x=2$ $h'(2)=16>0$ $h$ is increasing $\nearrow$ Now let's look at the critical points: $x$ Before After Verdict $-2$ $\nearrow$ $\searrow$ Maximum $0$ $\searrow$ $\searrow$ Not an extremum $1$ $\searrow$ $\nearrow$ Minimum Now we can see that $h$ has one relative minimum.